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Chapter 4: Problem 146

A 0.9157 -g mixture of \(\mathrm{CaBr}_{2}\) and \(\mathrm{NaBr}\) is dissolved in water, and \(\mathrm{AgNO}_{3}\) is added to the solution to form AgBr precipitate. If the mass of the precipitate is \(1.6930 \mathrm{~g}\), what is the percent by mass of \(\mathrm{NaBr}\) in the original mixture?

Short Answer

Expert verified
The percent by mass of \(\mathrm{NaBr}\) in the original mixture is approximately 55.56%.

Step by step solution

01

Calculate moles of AgBr

The molar mass of AgBr is \(331.232 \, \mathrm{g/mol}\). With a given mass of AgBr as \(1.6930 \, \mathrm{g}\), the number of moles can be calculated as follows: \[ \frac{1.6930 \, \mathrm{g}}{331.232 \, \mathrm{g/mol}} = 0.0051 \, \mathrm{mol}\]
02

Compute moles of Br in the original mixture

Each mole of AgBr has 1 mole of Br. Hence, the amount of Bromine in the original mixture is the same as in the precipitate, i.e., 0.0051 moles.
03

Calculate mass of Br in the original mixture

The molar mass of Br is \(79.904 \, \mathrm{g/mol}\). So, the mass of Bromine in the original mixture can be calculated as \[79.904 \, \mathrm{g/mol} \times 0.0051 \, \mathrm{mol} = 0.4071 \, \mathrm{g}\]
04

Compute mass of \(\mathrm{NaBr}\) in the original mixture

Since all of the Bromine originates from either \(\mathrm{CaBr}_{2}\) or \(\mathrm{NaBr}\), and the total mass of the mixture has been provided, we can figure out the mass of \(\mathrm{NaBr}\) in the original mixture. This is done by subtracting the mass of Bromine from the total mass of the mixture: \[0.9157 \, \mathrm{g} - 0.4071 \, \mathrm{g} = 0.5086 \, \mathrm{g}\]
05

Calculate percent by mass of \(\mathrm{NaBr}\) in the original mixture

Percent by mass is calculated as follows: \[ \frac{0.5086 \, \mathrm{g}}{0.9157 \, \mathrm{g}} \times 100 = 55.56 \% \]

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Most popular questions from this chapter

Oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) is present in many plants and vegetables. If \(24.0 \mathrm{~mL}\) of \(0.0100 \mathrm{M} \mathrm{KMnO}_{4}\) solution is needed to titrate \(1.00 \mathrm{~g}\) of a sample of \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) to the equivalence point, what is the percent by mass of \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) in the sample? The net ionic equation is \(2 \mathrm{MnO}_{4}^{-}+16 \mathrm{H}^{+}+5 \mathrm{C}_{2} \mathrm{O}_{4}^{2-} \longrightarrow\) \(2 \mathrm{Mn}^{2+}+10 \mathrm{CO}_{2}+8 \mathrm{H}_{2} \mathrm{O}\)

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